More answers

so is the directrix above or below the focus...? and then is the parabola concave up or down..?

campbell_st

2 years ago

|dw:1438892282167:dw|
so you plot looks like this

anonymous

2 years ago

below right?

campbell_st

2 years ago

|dw:1438892378795:dw|
the distance is labelled 2a and is double the focal length.

anonymous

2 years ago

8

campbell_st

2 years ago

the dotted line through x = 4 is the line of symmetry.... and the vertex is on this line midway between the focus and directrix

campbell_st

2 years ago

great so if the
twice the focal length is 8 then the focal length is a = 4
so the vertex is 4 units below the focus on the line x = 4
so where do you think the vertex is..?

campbell_st

2 years ago

|dw:1438892601545:dw|

anonymous

2 years ago

at y=-11 ??

anonymous

2 years ago

or no

campbell_st

2 years ago

great... so the standard form I use now is
\[(x - h)^2 = 4a(y - k)\]
(h, k) is the vertex and a is the focal length
so the equation is
\[(x -4)^2 = 4 \times 4(y + 11)\]
now you just need to simplify this equation

anonymous

2 years ago

f(x)=1/16x^-8x+11 ?

campbell_st

2 years ago

so if you want it in vertex form its
\[\frac{1}{16}(x -4)^2 = y + 11\]
then subtract 11 from both sides